3.601 \(\int \frac{(a+b x^2)^2 \sqrt{c+d x^2}}{x} \, dx\)

Optimal. Leaf size=92 \[ a^2 \sqrt{c+d x^2}-a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )-\frac{b \left (c+d x^2\right )^{3/2} (b c-2 a d)}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

[Out]

a^2*Sqrt[c + d*x^2] - (b*(b*c - 2*a*d)*(c + d*x^2)^(3/2))/(3*d^2) + (b^2*(c + d*x^2)^(5/2))/(5*d^2) - a^2*Sqrt
[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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Rubi [A]  time = 0.0841289, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {446, 88, 50, 63, 208} \[ a^2 \sqrt{c+d x^2}-a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )-\frac{b \left (c+d x^2\right )^{3/2} (b c-2 a d)}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x^2)^2*Sqrt[c + d*x^2])/x,x]

[Out]

a^2*Sqrt[c + d*x^2] - (b*(b*c - 2*a*d)*(c + d*x^2)^(3/2))/(3*d^2) + (b^2*(c + d*x^2)^(5/2))/(5*d^2) - a^2*Sqrt
[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^2 \sqrt{c+d x^2}}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(a+b x)^2 \sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (-\frac{b (b c-2 a d) \sqrt{c+d x}}{d}+\frac{a^2 \sqrt{c+d x}}{x}+\frac{b^2 (c+d x)^{3/2}}{d}\right ) \, dx,x,x^2\right )\\ &=-\frac{b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac{1}{2} a^2 \operatorname{Subst}\left (\int \frac{\sqrt{c+d x}}{x} \, dx,x,x^2\right )\\ &=a^2 \sqrt{c+d x^2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac{1}{2} \left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{c+d x}} \, dx,x,x^2\right )\\ &=a^2 \sqrt{c+d x^2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}+\frac{\left (a^2 c\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{d}+\frac{x^2}{d}} \, dx,x,\sqrt{c+d x^2}\right )}{d}\\ &=a^2 \sqrt{c+d x^2}-\frac{b (b c-2 a d) \left (c+d x^2\right )^{3/2}}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2}-a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )\\ \end{align*}

Mathematica [A]  time = 0.098136, size = 93, normalized size = 1.01 \[ a^2 \sqrt{c+d x^2}-a^2 \sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c+d x^2}}{\sqrt{c}}\right )+\frac{b \left (c+d x^2\right )^{3/2} (2 a d-b c)}{3 d^2}+\frac{b^2 \left (c+d x^2\right )^{5/2}}{5 d^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x^2)^2*Sqrt[c + d*x^2])/x,x]

[Out]

a^2*Sqrt[c + d*x^2] + (b*(-(b*c) + 2*a*d)*(c + d*x^2)^(3/2))/(3*d^2) + (b^2*(c + d*x^2)^(5/2))/(5*d^2) - a^2*S
qrt[c]*ArcTanh[Sqrt[c + d*x^2]/Sqrt[c]]

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Maple [A]  time = 0.01, size = 100, normalized size = 1.1 \begin{align*}{\frac{{b}^{2}{x}^{2}}{5\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-{\frac{2\,{b}^{2}c}{15\,{d}^{2}} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}+{\frac{2\,ab}{3\,d} \left ( d{x}^{2}+c \right ) ^{{\frac{3}{2}}}}-\sqrt{c}\ln \left ({\frac{1}{x} \left ( 2\,c+2\,\sqrt{c}\sqrt{d{x}^{2}+c} \right ) } \right ){a}^{2}+{a}^{2}\sqrt{d{x}^{2}+c} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x)

[Out]

1/5*b^2*x^2*(d*x^2+c)^(3/2)/d-2/15*b^2*c/d^2*(d*x^2+c)^(3/2)+2/3*a*b*(d*x^2+c)^(3/2)/d-c^(1/2)*ln((2*c+2*c^(1/
2)*(d*x^2+c)^(1/2))/x)*a^2+a^2*(d*x^2+c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.65494, size = 481, normalized size = 5.23 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{c} d^{2} \log \left (-\frac{d x^{2} - 2 \, \sqrt{d x^{2} + c} \sqrt{c} + 2 \, c}{x^{2}}\right ) + 2 \,{\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} +{\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{30 \, d^{2}}, \frac{15 \, a^{2} \sqrt{-c} d^{2} \arctan \left (\frac{\sqrt{-c}}{\sqrt{d x^{2} + c}}\right ) +{\left (3 \, b^{2} d^{2} x^{4} - 2 \, b^{2} c^{2} + 10 \, a b c d + 15 \, a^{2} d^{2} +{\left (b^{2} c d + 10 \, a b d^{2}\right )} x^{2}\right )} \sqrt{d x^{2} + c}}{15 \, d^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="fricas")

[Out]

[1/30*(15*a^2*sqrt(c)*d^2*log(-(d*x^2 - 2*sqrt(d*x^2 + c)*sqrt(c) + 2*c)/x^2) + 2*(3*b^2*d^2*x^4 - 2*b^2*c^2 +
10*a*b*c*d + 15*a^2*d^2 + (b^2*c*d + 10*a*b*d^2)*x^2)*sqrt(d*x^2 + c))/d^2, 1/15*(15*a^2*sqrt(-c)*d^2*arctan(
sqrt(-c)/sqrt(d*x^2 + c)) + (3*b^2*d^2*x^4 - 2*b^2*c^2 + 10*a*b*c*d + 15*a^2*d^2 + (b^2*c*d + 10*a*b*d^2)*x^2)
*sqrt(d*x^2 + c))/d^2]

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Sympy [A]  time = 25.1065, size = 90, normalized size = 0.98 \begin{align*} \frac{a^{2} c \operatorname{atan}{\left (\frac{\sqrt{c + d x^{2}}}{\sqrt{- c}} \right )}}{\sqrt{- c}} + a^{2} \sqrt{c + d x^{2}} + \frac{b^{2} \left (c + d x^{2}\right )^{\frac{5}{2}}}{5 d^{2}} + \frac{\left (c + d x^{2}\right )^{\frac{3}{2}} \left (4 a b d - 2 b^{2} c\right )}{6 d^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**2*(d*x**2+c)**(1/2)/x,x)

[Out]

a**2*c*atan(sqrt(c + d*x**2)/sqrt(-c))/sqrt(-c) + a**2*sqrt(c + d*x**2) + b**2*(c + d*x**2)**(5/2)/(5*d**2) +
(c + d*x**2)**(3/2)*(4*a*b*d - 2*b**2*c)/(6*d**2)

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Giac [A]  time = 1.12601, size = 136, normalized size = 1.48 \begin{align*} \frac{a^{2} c \arctan \left (\frac{\sqrt{d x^{2} + c}}{\sqrt{-c}}\right )}{\sqrt{-c}} + \frac{3 \,{\left (d x^{2} + c\right )}^{\frac{5}{2}} b^{2} d^{8} - 5 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} b^{2} c d^{8} + 10 \,{\left (d x^{2} + c\right )}^{\frac{3}{2}} a b d^{9} + 15 \, \sqrt{d x^{2} + c} a^{2} d^{10}}{15 \, d^{10}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^2*(d*x^2+c)^(1/2)/x,x, algorithm="giac")

[Out]

a^2*c*arctan(sqrt(d*x^2 + c)/sqrt(-c))/sqrt(-c) + 1/15*(3*(d*x^2 + c)^(5/2)*b^2*d^8 - 5*(d*x^2 + c)^(3/2)*b^2*
c*d^8 + 10*(d*x^2 + c)^(3/2)*a*b*d^9 + 15*sqrt(d*x^2 + c)*a^2*d^10)/d^10